At release,
$$
C=\frac{\mu}{2a_p}-\frac{\mu(1-\beta)}{a_p}
=\frac{\mu}{a_p}\left(\beta-\frac12\right).
$$
At large radius the potential vanishes, so
$$
\boxed{v_\infty=\sqrt{2C}
=\sqrt{(2\beta-1)\frac{\mu}{a_p}}
\simeq\sqrt{\frac{2\beta\mu}{a_p}}}.
$$
The asymptote satisfies $1+e\cos f_\infty=0$, where $f=\theta-\varpi$. If the angular displacement from the release direction is $\Delta=\pi-f_\infty$, then
$$
\cos(\pi-\Delta)=-\frac1e=-1+\frac1\beta.
$$
For $\beta\gg1$, $-\cos\Delta\simeq-1+\Delta^2/2$, and therefore
$$
\boxed{\Delta\simeq\sqrt{\frac2\beta}}.
$$


Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 316 / 2 / iv / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/iii/paper-316/2/iv/solution